Answer:
62.17% probability that a randomly selected exam will require more than 15 minutes to grade
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean
and standard deviation
, the zscore of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:

What is the probability that a randomly selected exam will require more than 15 minutes to grade
This is 1 subtracted by the pvalue of Z when X = 15. So



has a pvalue of 0.3783.
1 - 0.3783 = 0.6217
62.17% probability that a randomly selected exam will require more than 15 minutes to grade
Answer: 96 adult tickets were sold.
Step-by-step explanation:
Let x represent the number of adult tickets that were sold.
Let y represent the number of students tickets that were sold.
For the last basketball game, 381 tickets were sold. This means that
x + y = 381
Adult tickets sold for $7, and student tickets sold for $3. If the ticket sales totaled 1,527, it means that
7x + 3y = 1527 - - - - - - - - - - - -1
Substituting x = 381 - y into equation 1, it becomes
7(381 - y) + 3y = 1527
2667 - 7y + 3y = 1527
- 7y + 3y = 1527 - 2667
- 4y = - 1140
y = - 1140/- 4
y = 285
x = 381 - y = 381 - 285
x = 96
Answer:
The sum of two negative integers is always negative.
Step-by-step explanation:
For example, add -1+-1 which =-2, which are the smallest neggative numbers possible.-100+-100=-200, and -4+-4=-8, so you see that the sum of two negative integers should always be negative.
0.83Answer:
Step-by-step explanation:
83% means 83 per 100 or 83/100 if you divide 83 by 100 you get 0.83
Answer:
48.225 ft squared
Step-by-step explanation: